# 11. Container With Most Water

Reference: LeetCode

Difficulty: Medium

My Post: [Java] Two Pointers with Explanation (easy-understand)

## Problem

Given

`n`

non-negativeintegers`a1, a2, ..., an`

, where each represents a point at coordinate`(i, ai)`

.`n`

vertical lines are drawn such that the two endpoints of line`i`

is at`(i, ai)`

and`(i, 0)`

. Find two lines, which together with x-axis forms a container, such that the container contains the most water.

**Note:** You may not slant the container and `n`

is at least 2.

**Example:**

1 | Input: [1,8,6,2,5,4,8,3,7] |

## Analysis

### Brute-Force

Try every container.

1 | public int maxArea(int[] height) { |

**Time:** $O(N^2)$**Space:** $O(1)$

### Two Pointers

We have two pointers `lo`

and `hi`

that start from two ends. Each time we update the one with smaller height.

**Why?**

It is kind of pruning unnecessary cases. If `height[lo] < height[hi]`

, we move `lo`

because we know that the areas in which `lo`

is the left bar (`[lo, lo + 1]`

,`[lo, lo + 2]`

, …, `[lo, hi - 1]`

) must be **less than** the area in `[lo, hi]`

that we just considered. So next time we should not consider `lo`

as the left end anymore.

For example, all the areas (`AB`

, `AC`

, `AD`

, …, `AH`

) are less than `AI`

.

1 | public int maxArea(int[] height) { |

**Time:** $O(N)$**Space:** $O(1)$